The ratio of the amount of two element X and Y at radioactive equilibrium is `1 : 2 xx 10^(-6)`. If the half-life period of element Y is `4.9 xx 10^(-4)` days, then the half-life period of element X will be

Options

A. `4.8 xx 10^(-3)` days
B. 245 days
C. 122.5 days
D. None of these

Explanation

To solve the problem, we need to determine the half-life period of element X given the half-life of element Y and the ratio of their amounts at radioactive equilibrium.

### Step-by-Step Solution:

1. **Understand the Concept of Radioactive Equilibrium**:
In radioactive equilibrium, the decay rate (or activity) of the parent isotope (X) equals the decay rate of the daughter isotope (Y). This can be expressed as:
\[ \lambda_X \cdot N_X = \lambda_Y \cdot N_Y \]
where \( \lambda \) is the decay constant and \( N \) is the amount of the isotope.

2. **Express Decay Constants in Terms of Half-Lives**:
The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Therefore, we can express the decay constants for X and Y as:
\[ \lambda_X = \frac{0.693}{t_{1/2}^X} \quad \text{and} \quad \lambda_Y = \frac{0.693}{t_{1/2}^Y} \]

3. **Set Up the Ratio Based on Given Information**:
We know the ratio of the amounts of X and Y is given as:
\[ \frac{N_X}{N_Y} = 1 : 2 \times 10^{-6} \]
This can be rewritten as:
\[ N_X = 1 \quad \text{and} \quad N_Y = 2 \times 10^{-6} \]

4. **Substitute into the Equilibrium Equation**:
Substituting the amounts into the equilibrium equation gives:
\[ \frac{0.693}{t_{1/2}^X} \cdot 1 = \frac{0.693}{t_{1/2}^Y} \cdot (2 \times 10^{-6}) \]

5. **Simplify the Equation**:
Canceling \( 0.693 \) from both sides, we have:
\[ \frac{1}{t_{1/2}^X} = \frac{2 \times 10^{-6}}{t_{1/2}^Y} \]

6. **Substituting the Known Half-Life of Y**:
The half-life of Y is given as:
\[ t_{1/2}^Y = 4.9 \times 10^{-4} \text{ days} \]
Substituting this into the equation gives:
\[ \frac{1}{t_{1/2}^X} = \frac{2 \times 10^{-6}}{4.9 \times 10^{-4}} \]

7. **Calculate \( t_{1/2}^X \)**:
Rearranging gives:
\[ t_{1/2}^X = \frac{4.9 \times 10^{-4}}{2 \times 10^{-6}} = \frac{4.9 \times 10^{-4}}{2} \times 10^{6} \]
This simplifies to:
\[ t_{1/2}^X = 2.45 \times 10^{2} \text{ days} = 245 \text{ days} \]

### Final Answer:
The half-life period of element X is **245 days**.